The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102...

1. Translate the problem requirements: We need to find the sum of even integers from 102 to 200 inclusive. We're given that the sum of the first 50 positive even integers \(2, 4, 6, \ldots, 100\) equals 2550.
2. Identify the sequence structure: Determine which even integers we're dealing with in our target range and how they relate to the first 50 even integers.
3. Connect to the given information: Use the fact that we know the sum of the first 50 even integers to find a pattern or relationship that helps us calculate our target sum.
4. Apply the relationship: Calculate the sum of even integers from 102 to 200 using the connection we established.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're given and what we need to find.

We're told that the sum of the first 50 positive even integers equals 2550. The first 50 positive even integers are: \(2, 4, 6, 8, \ldots, 100\). We can verify this makes sense - the 50th even integer is \(2 \times 50 = 100\).

We need to find the sum of even integers from 102 to 200 inclusive. Let's think about this range: it starts at 102 and goes up to 200, hitting every even number along the way \(102, 104, 106, \ldots, 200\).

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding of what sequence we're working with.

2. Identify the sequence structure

Now let's figure out exactly which even integers we're dealing with and how many there are.

The even integers from 102 to 200 are: \(102, 104, 106, 108, \ldots, 198, 200\).

To count how many terms this is, let's think about it step by step:

• The first term is 102
• The last term is 200
• Each term increases by 2

To find the number of terms: \((200 - 102) \div 2 + 1 = 98 \div 2 + 1 = 49 + 1 = 50\) terms.

Interesting! We have exactly 50 terms here too, just like in the given information.

Now, let's see how these terms relate to the first 50 even integers. The first 50 even integers are \(2, 4, 6, \ldots, 100\). Our sequence is \(102, 104, 106, \ldots, 200\). Notice that each term in our sequence is exactly 100 more than the corresponding term in the first sequence:

• \(102 = 2 + 100\)
• \(104 = 4 + 100\)
• \(106 = 6 + 100\)
• And so on...

3. Connect to the given information

Since each term in our target sequence (102 to 200) is exactly 100 more than the corresponding term in the known sequence (2 to 100), we can use this relationship to find our answer.

The sum of our sequence = \((2 + 100) + (4 + 100) + (6 + 100) + \ldots + (100 + 100)\)

We can rearrange this as:
Sum = \((2 + 4 + 6 + \ldots + 100) + (100 + 100 + 100 + \ldots + 100)\)

The first part \((2 + 4 + 6 + \ldots + 100)\) is given as 2550.

The second part is 100 added 50 times (since we have 50 terms), which equals \(100 \times 50 = 5000\).

Process Skill: INFER - Drawing the non-obvious conclusion that we can split the sum and use the given information.

4. Apply the relationship

Now we can calculate our final answer:

Sum of even integers from 102 to 200 = Sum of first 50 even integers + \((100 \times 50)\)
Sum = \(2550 + 5000 = 7550\)

Let's verify this makes sense by checking our logic:

• We have 50 terms in our sequence
• Each term is 100 more than the corresponding term in the known sequence
• So our sum should be \(100 \times 50 = 5000\) more than the known sum
• \(2550 + 5000 = 7550\) ✓

Final Answer

The sum of even integers from 102 to 200 inclusive is 7,550.

This matches answer choice B. 7,550.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the range "from 102 to 200 inclusive"
Students may incorrectly interpret this as starting from 100 or ending at 198, missing that "inclusive" means both 102 and 200 are included in the sequence. This leads to counting the wrong number of terms.

2. Failing to recognize the relationship between sequences
Students might try to use complex arithmetic sequence formulas instead of noticing that each term in the target sequence \((102, 104, 106\ldots)\) is exactly 100 more than the corresponding term in the given sequence \((2, 4, 6\ldots)\). Missing this pattern makes the problem much harder than necessary.

3. Incorrectly counting the number of terms
When calculating how many even integers are from 102 to 200, students often forget to add 1 to their calculation. They might compute \((200-102) \div 2 = 49\) and stop there, missing that the correct count is \(49+1 = 50\) terms.

Errors while executing the approach

1. Arithmetic errors in the final addition
Even with the correct approach, students may make simple calculation mistakes when computing \(2550 + 5000\), potentially getting 7050 or 8550 instead of 7550.

2. Incorrectly calculating 100 × 50
Students might miscalculate this basic multiplication, getting 500 or 5500 instead of 5000, which throws off the entire final answer.

Errors while selecting the answer

No likely faltering points - the calculation directly yields 7550, which clearly matches answer choice B.